Are meromorphic functions bounded "near" a pole?

Question

I am trying to understand the concept of a removeable singularity.

Number three says

"There exists a neighbourhood of $a$ on which $f$ is bounded"

This is confusing me since to me a neighbourhood is just a set containing an open set w.o reference to size.

A pole for instance is not bounded arbitraraly close to it, yet poles are reamovable.

Does anyone have such an example or can explain where my reasoning fails.

Answer 1

The assertion “There exists a neighbourhood of $a$ on which $f$ is bounded” means that there exists a neighbourhood $N$ of $a$ such that the set $f(N)=\{f(z)\,|\,z\in N\cap\text{domain of }f\}$ is bounded.

For instance, if $f\colon\mathbb{C}\setminus\{0\}\longrightarrow\mathbb C$ is defined by $f(z)=z$, then $0$ is a removable singularity and $f\bigl(D(0,1)\bigr)=D(0,1)\setminus\{0\}$, which is bounded.

By the way, poles are not removable singularities.

Answer 2

Just an addition to Jose Carlos Santos answer, the classification of singularities goes like this:

1. Removable (aka good) singularities are points where the function is bounded near them,
2. Poles (aka almost good) are points where the limit of the function is $\infty$ and the last one are
3. essential singularities (extremely bad) which are points that don't fit to one of the previous cases.

There is a great exposition by Terence Tao here about singularities where he explains in particular how the first two singularities can be "fixed".